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3如图所示，运动员把铅球从A点斜向上抛出，B点是铅球运动轨迹的最高点，C点是铅球下
2025年7月济南市高一期末学习质量检测
落过程中经过的某一点，不计空气阻力，下列说法正确的是
A铅球运动到B点时的速度为零
物理试题
B铅球从A点运动到B点的过程中处于失重状态
C.铅球在B点所受的合力比在C点所受的合力大
D.铅球在空中运动过程中的加速度方向一直在改变
本试卷分选择题和非选择题两部分，共100分。考试时间90分钟。
4.甲、乙两位同学研究运动的合成规律，如图所示，在一张白纸上以0点为4少
原点建立xOy坐标系，甲同学手章铅笔，让笔尖从O点由静止开始，紧贴
注意事项：
直尺沿x轴正方向匀加速画线，同时，乙同学推动直尺紧贴纸面沿着y
1答题前，考生将自己的姓名、考生号、座号填写在相应位置，认其核对条形码上的姓名、
轴正方向匀速运动，该过程中直尺始终保持与x轴平行，铅笔在白纸上留
考生号和座号，并将条形码粘贴在指定位置上
下的痕迹可能是
2.选择题答案必须使用2B铅笔（按填涂样例）正确填涂；非选择题答案必须使用0.5惑米
黑色签字笔书写，字体工整、笔迹清楚
3.请按照愿号在答题卡各题目的答题区域内作答，超出答题区城书写的答案无效：在草將
纸、试题卷上答题无效。保持卡面清洁，不折叠、不玻损。
5,如图所示为球心为O的球而。过球心O有一圆形截面DMCN,CD和MN是该圆形截面
上的两条直径，AB是球的某一直径且与圆形战面DMCN垂直。在A,B两点放置两个等
一、单项选择题〔本题共10个小题，每小题3分，共30分。每小题只有一个选项符合题目
量的异种点电荷，下列说法正确的是
要求)
A.M、N两点的电场强度不同
1.物体做匀速圆周运动的过程中，下列物理量发生变化的是
B.O点的电场强度为零
A加速度
B速率
C.角速度
D周期
C沿直径AB从A到B,电场强度先减小后增大
2如图所示，赛车手驾驶摩托车在水平路面上转弯时车身向内侧颀斜一定角度，在摩托车转弯
D.沿直径CD从C到D,电场强度先减小后增大
过程中，下列说法正确的是
6.如图所示，粗糙的固定斜面上，轻质弹簧一端栓接斜面底部的挡板，另一端栓接在物块上。
A地而对车轮的支持力垂直于水平路面向上
用外力作用在物块上，将弹簧毁慢压缩至A点后，撤去外力，物块由静止开始沿斜面向上运
B.地面对车轮的支持力沿车身的方向斜向上
动到最高点B后又沿斜面下滑。已知O点为弹资的原长位置，下列说法正确的是
℃如果摩托车发生侧带是因为赛车手与摩托车整体受到向外的
A物块下滑时能悠回到A点
B
力作用
B物块第一次向上经过O点的速度大于第一次向下经过O点的速度
C,物块沿斜面向上运动过程中运动至O点时动能最大
D赛车手与摩托车整体受到重力、支持力、摩擦力和向心力的作用
D.物块沿斜面向上运动过程中弹簧弹性势能的政变量等于物块机械能的改变量
高一物理试题第1页（共8页）
高一物理试题第2页（共8页）

Q夸克扫描王
极速扫描，就是高效可2025年 7月济南市高一期末学习质量检测
参考答案
题号 1 2 3 4 5 6 7 8 9 10
答案 A A B D C B C D D C
题号 11 12 13 14 15 16
答案 AB AD BD BC ACD CD
17.（6分）(1) B (2) C (3) 大于 （每空 2分）
d
18. 6 1 rd
2 rbd 2 2
（ 分）（ ） （每空 1分） (2) aR （每空 2分）
Rt R2t 2 gR2 rbd 2
19.（7分）
（1）加速阶段：v2=2aS ································································································································1分
μmgcosθ－mgsinθ= ma········································································· 1分
解得：a=0.4m/s2
μ=0.8 ···················································································1 分
（2）Q=μmgcosθ(vt-S) ···················································································· 2 分
v
t= ························································································ 1 分
a
解得：Q=32J ·····························································································1 分
20.（7分）
1
（1）A到 B：H -h gt 21 ··············································································1 分2
l=v0cos45°t1 ··········································································· 1 分
v0sin45°=gt1 ············································································· 1 分
解得 t1=2s，l=40m ···········································································1 分
（2）B到 C：S=vBcosαt2 ··················································································································1 分
gt2=2vBsinα ·········································································1 分
v 2B sin 2 解得 S=
g
当α=45°时，B点最小速度 vB=20m/s ····························································1 分
{#{QQABJQ6l5wg4kAZACA76F0H8CgmQkIKSLYoGgQAWuAYKSAFIFAA=}#}
21.（9分）
（1）mgsinθ－qE=ma ·············································································· 2 分
1
解得：a= gsinθ ············································································ 1 分
2
（2）mgLsinθ－W=Ek ············································································· 1 分
W q EO E L L ·················································································1 分
2
3
解得：Ek= mgLsinθ ···········································································1 分
8
（3）mgxsinθ－ΔEp=0 ··········································································1 分

q 1 mg sin mg sin mg sin x

ΔEp＝ x ·································· 1 分2 2q 2q 4qL
解得：x=4L，ΔEp=4mgLsinθ ································································· 1 分
22.（11分）
v 2
（1）在 C：mg=m C ····················································································1 分
R
1
从弹射到 C：Ep1=2mgR+ mv 2C ······························································2 分2
Ep1=3J ················································································ 1 分
（2）若物块恰好到达与圆心等高处：Ep0=mgR=1.2J ········································ 1 分
若物块恰好通过竖直轨道最高点：Ep1=3J
若物块恰好到达 E点：Ep2=μ1mgDE=4.8J ····················································1 分
若物块恰好到达平板的右端： 1mg 2 M m g Ma ·····························1 分
v gt at L v t 1 gt 2 1E 1 E 1 at
2 ···········································1 分
2 2
解得 vE= 2 2 m/s
1
E 2p3－μ1mgs= mv ··············································································· 1 分
2 E
解得 Ep3=6J ···············································································1 分
综上：0≤Ep≤1.2J，或 3J≤Ep≤4.8J，或 Ep>6J ··········································· 1 分
{#{QQABJQ6l5wg4kAZACA76F0H8CgmQkIKSLYoGgQAWuAYKSAFIFAA=}#}

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