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3如圖所示，運動員把鉛球從A點斜向上拋出，B點是鉛球運動軌跡的最高點，C點是鉛球下
2025年7月濟南市高一期末學(xué)習(xí)質(zhì)量檢測
落過程中經(jīng)過的某一點，不計空氣阻力，下列說法正確的是
A鉛球運動到B點時的速度為零
物理試題
B鉛球從A點運動到B點的過程中處于失重狀態(tài)
C.鉛球在B點所受的合力比在C點所受的合力大
D.鉛球在空中運動過程中的加速度方向一直在改變
本試卷分選擇題和非選擇題兩部分，共100分。考試時間90分鐘。
4.甲、乙兩位同學(xué)研究運動的合成規(guī)律，如圖所示，在一張白紙上以0點為4少
原點建立xOy坐標(biāo)系，甲同學(xué)手章鉛筆，讓筆尖從O點由靜止開始，緊貼
注意事項：
直尺沿x軸正方向勻加速畫線，同時，乙同學(xué)推動直尺緊貼紙面沿著y
1答題前，考生將自己的姓名、考生號、座號填寫在相應(yīng)位置，認(rèn)其核對條形碼上的姓名、
軸正方向勻速運動，該過程中直尺始終保持與x軸平行，鉛筆在白紙上留
考生號和座號，并將條形碼粘貼在指定位置上
下的痕跡可能是
2.選擇題答案必須使用2B鉛筆（按填涂樣例）正確填涂；非選擇題答案必須使用0.5惑米
黑色簽字筆書寫，字體工整、筆跡清楚
3.請按照愿號在答題卡各題目的答題區(qū)域內(nèi)作答，超出答題區(qū)城書寫的答案無效：在草將
紙、試題卷上答題無效。保持卡面清潔，不折疊、不玻損。
5,如圖所示為球心為O的球而。過球心O有一圓形截面DMCN,CD和MN是該圓形截面
上的兩條直徑，AB是球的某一直徑且與圓形戰(zhàn)面DMCN垂直。在A,B兩點放置兩個等
一、單項選擇題〔本題共10個小題，每小題3分，共30分。每小題只有一個選項符合題目
量的異種點電荷，下列說法正確的是
要求)
A.M、N兩點的電場強度不同
1.物體做勻速圓周運動的過程中，下列物理量發(fā)生變化的是
B.O點的電場強度為零
A加速度
B速率
C.角速度
D周期
C沿直徑AB從A到B,電場強度先減小后增大
2如圖所示，賽車手駕駛摩托車在水平路面上轉(zhuǎn)彎時車身向內(nèi)側(cè)頎斜一定角度，在摩托車轉(zhuǎn)彎
D.沿直徑CD從C到D,電場強度先減小后增大
過程中，下列說法正確的是
6.如圖所示，粗糙的固定斜面上，輕質(zhì)彈簧一端栓接斜面底部的擋板，另一端栓接在物塊上。
A地而對車輪的支持力垂直于水平路面向上
用外力作用在物塊上，將彈簧毀慢壓縮至A點后，撤去外力，物塊由靜止開始沿斜面向上運
B.地面對車輪的支持力沿車身的方向斜向上
動到最高點B后又沿斜面下滑。已知O點為彈資的原長位置，下列說法正確的是
℃如果摩托車發(fā)生側(cè)帶是因為賽車手與摩托車整體受到向外的
A物塊下滑時能悠回到A點
B
力作用
B物塊第一次向上經(jīng)過O點的速度大于第一次向下經(jīng)過O點的速度
C,物塊沿斜面向上運動過程中運動至O點時動能最大
D賽車手與摩托車整體受到重力、支持力、摩擦力和向心力的作用
D.物塊沿斜面向上運動過程中彈簧彈性勢能的政變量等于物塊機械能的改變量
高一物理試題第1頁（共8頁）
高一物理試題第2頁（共8頁）

Q夸克掃描王
極速掃描，就是高效可2025年 7月濟南市高一期末學(xué)習(xí)質(zhì)量檢測
參考答案
題號 1 2 3 4 5 6 7 8 9 10
答案 A A B D C B C D D C
題號 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）加速階段：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
當(dāng)α=45°時，B點最小速度 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
從彈射到 C：Ep1=2mgR+ mv 2C ······························································2 分2
Ep1=3J ················································································ 1 分
（2）若物塊恰好到達與圓心等高處：Ep0=mgR=1.2J ········································ 1 分
若物塊恰好通過豎直軌道最高點：Ep1=3J
若物塊恰好到達 E點：Ep2=μ1mgDE=4.8J ····················································1 分
若物塊恰好到達平板的右端： 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
綜上：0≤Ep≤1.2J，或 3J≤Ep≤4.8J，或 Ep>6J ··········································· 1 分
{#{QQABJQ6l5wg4kAZACA76F0H8CgmQkIKSLYoGgQAWuAYKSAFIFAA=}#}

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