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子江中学2017-2018学年高三期初考试

子江中学2017-2018学年高三期初考试
子江中学2017-2018学年高三期初考试

子江中学2017-2018学年高三期初物理试卷2017.8.30

一、单选题(共5小题,共15分)

1.关于质点,下列说法正确的是()

A.体积很小的物体都可以看作质点

B.研究花样滑冰运动员优美的动作时,可以把运动员看作质点

C.研究一列火车通过长江大桥所需的时间内,可以把火车看作质点

D.物体的大小和形状在所研究的问题中起的作用很小,可以忽略不计时,我们就可以把物体看作质点

2.如图所示,从斜面上某一位置先后由静止释放四个小球.相邻两小球

释放的时间间隔为0.1s,某时刻拍下小球所处位置的照片,测出

x AB=5cm,x BC=10cm,x CD=15cm.则()

A.小球从A点释放

B.C点小球速度是A、D点小球速度之和的一半

C.B点小球的速度大小为1.5m/s

D.小球的加速度大小为5m/s2

3.某中学生正在水平地面上试验自制的一枚火箭,火箭经点

火升空后,在竖直方向上运动的v-t图象如图所示,则下列

说法中正确的是()

A.0~1s内火箭匀速上升

B.1~2s内火箭静止不动

C.3s末火箭上升到最高点

D.5s末火箭上升到最高点

4.木块A、B分别重50N和30N,它们与水平地面之间的动摩擦因

数均为0.2.与A、B相连接的轻弹簧被压缩了5cm,系统置于水

平地面上静止不动.已知弹簧的劲度系数为100N/m.用F=2N的

不变力作用在木块A上,如图所示.力F作用后()

A.木块A所受摩擦力大小是9N,方向向右

B.木块A所受摩擦力大小是3N,方向向右

C.木块B所受摩擦力大小是9N,方向向左

D.木块B所受摩擦力大小是3N,方向向左

5.如图所示,A、B两物体通过轻细绳跨过定滑轮相连接,已知物体A的质量大

于物体B的质量,开始它们处于静止状态.在水平拉力F的作用下,物体A向

右做变速运动,同时物体B匀速上升.设水平地面对物体A的支持力为N,对A

的摩擦力为f,绳子对A的拉力为T.那么在物体B匀速上升的过程中,下列说

法中正确的是()

A. N、f增大,T不变

B.N、f减小、T不变

C. N、f、T都增大

D.N、T减小,f不变

二、多选题(本题共5小题,共15分)

6. 做匀加速直线运动的物体的加速度为3m/s2,对任意1s来说,下列说法中正确的是()

A.某1s末的速度比该1s初的速度总是大3m/s

B.某1s末的速度比前1s初的速度大3m/s

C.某1s末的速度比前1s末的速度大3m/s

D.某1s末的速度比该1s初的速度总是大3倍

7.物体从h高处自由下落,它在落地前1s内共下落35m,g=10m/s2.下列说法中正确的有()

A.物体下落的时间为4s

B.物体落地时速度为40m/s

C.下落后第1s内、第2s内、第3s内,每段位移之比为1:2:3

D.前2s内共下落60m

8.如图甲所示,水平传送带始终以恒定

速率υ1向右运行.质量为m的物块,

以υ2的初速度从与传送带等高的光滑

水平地面上的A处向左滑入传送带.若

从物块滑上传送带开始计时,物块在传

送带上运动的υ-t图象(以地面为参考系)如图乙所示.已知υ2>υ1,则()

A.t1时刻,物块离A处的距离达到最大

B.t2时刻,物块相对传送带滑动的距离达到最大

C.0-t3时间内,物块受到的摩擦力方向一直向右

D.t1-t2时间内,物块做匀加速运动

9.如图所示,水平细杆上套一细环A,环A与球B间用一轻

质绳相连,质量分别为m A、m B,由于B球受到水平恒力F的作

用,A环与B球一起向右匀速运动,绳与竖直方向的夹角为

θ.则下列说法正确的是()

A.B球受到的恒力F大小为m A gtanθ

B.A环与杆的动摩擦因数为tanθ

C.力F逐渐增大,杆对A环的弹力保持不变

D.力F逐渐增大,绳对B球的拉力保持不变

10.如图所示,在光滑水平桌面的两端各固定一个等高的定滑

轮,用轻绳经过滑轮将测力计分别与质量为m1=8kg、m2=2kg的

两个物体A、B相连.不计绳与滑轮之间的摩擦及测力计的质量,

取g=10m/s2.放手后两物体开始运动,在两物体运动过程中

()

A.物体B的加速度大小为8m/s2

B.物体A的加速度大小为6m/s2

C. 测力计的读数为60N

D. 测力计的读数为32N

三、实验题探究题(本大题共2小题,共12.0分)

11.某同学在做“验证力的平行四边形定则”的实验中,主要实验步骤如下:

A.在桌面上放一块木板,在木板上铺一张白纸,用图钉把白纸钉在木板上

B.用图钉把橡皮条的一端固定在木板上的A点,在橡皮条的另一端拴上两条细绳,细

绳的另一端打成绳套

C.用两个弹簧秤分别勾住绳套,平行于木板且互成角度地拉橡皮条,把橡皮条的结点拉到某一位置O,记录下O点的位置,读出两个弹簧秤的示数

D.按选好的比例,用铅笔和刻度尺作出两个弹簧秤的拉力F1和F2的图示,并用平行四边形定则求出合力F′

E.只用一个弹簧秤,通过细绳套拉橡皮条使其伸长,读出弹簧秤的示数,记下细绳的方向,按同一比例作出这个力F的图示

F.比较力F′和F的大小和方向,看它们是否相同,得出结论

(1)上述步骤中:

①有重要遗漏的步骤序号是 ______ 和 ______ ;

②遗漏的内容是 _____ _ .

(2)某同学的实验情况如图甲所示,其中A为固定橡皮条的图钉,O为橡皮条与细绳的结点,OB和OC为细绳.

①图乙是在白纸上根据实验结果画出的力的图示,下列说法中正确的是 ______

A.图乙中的F是力F1和F2合力的理论值,F′是力F1和F2合力的实际测量值

B.图乙的F′是力F1和F2合力的理论值,F是力F1和F2合力的实际测量值

C.在实验中,如果将细绳也换成橡皮条,那么对实验结果没有影响

D.在实验中,如果将细绳也换成橡皮条,那么对实验结果有影响

②本实验采用的科学方法是 ______ (填字母代号)

A.理想实验法B.等效替代法

C.控制变量法D.建立物理模型法

③在“验证力的平行四边形定则”中,采取下列哪些方法和步骤可减小实验误差 ______ A.两个分力F1、F2间的夹角越大越好

B.两个分力F1、F2的大小在橡皮条的弹性限度内要适当大些

C.拉橡皮条的细绳要稍长一些,标记同一细绳方向的两个点要近些

D.实验前先把两个弹簧秤的钩子互相钩住,平放在桌子上,向相反方向拉动,检查读数是否相同

E.测力计、细绳、橡皮条都应与木板平行.

12.图为“测量匀变速直线运动的加速度”的实验装置示意图.实验步骤如下:

①测得遮光片的宽度为b;两光电门之间的距离为d;

②调整轻滑轮,使细线水平;

③让物块从光电门A的左侧由静止释放,用数字毫秒计分别测出遮光片经过光电门A和光电门B所用的时间△t1和△t2,求出加速度a;

④多次重复步骤③,求a的平均值;

回答下列为题:

(1)物块的加速度a可用d、b、△t1,和△t2,表示为a= ______

(2)(多选题)为减小实验误差,可采取的方法是 ______

四、计算题(本题共3小题,共8+8+10=26分)

13.一辆长途客车正在以20m/s的速度匀速行驶.突然,司机看见

车的正前方33m处有一只狗,如图所示,司机立即采取制动措

施.若从司机看见狗开始计时(t=0),长途客车司机的反应时间为

0.5s,制动过程中加速度大小为5m/s2,g取l0m/s2.

(1)求长途客车从司机发现狗至停止运动的这段时间内前进的距离.

(2)若狗正以4m/s的速度与长途客车同向奔跑,问狗能否摆脱被撞的噩运?

14.重为G1=8N的砝码悬挂在绳PA和PB的结点P上.PA偏离竖直

方向37°角,PB沿水平方向,且与重为G2=100N的木块相连接,

木块静止于倾角为37°的斜面上,如图所示,试求:(sin37°=0.6,

cos37°=0.8)

(1)木块与斜面间的摩擦力大小;

(2)斜面对木块的弹力大小.

15.如图所示,质量M=8kg的小车放在水平光滑的平面上,在小车左端加一水平推力F=8N,

当小车向右运动的速度达到1.5m/s时,在小车前端轻轻地放

上一个大小不计,质量为m=2kg的小物块,物块与小车间的动摩擦因数μ=0.2,小车足够长(取g=l0m/s2).求:

(1)小物块放后,小物块及小车的加速度大小各为多大?

(2)经多长时间两者达到相同的速度?

(3)从小物块放上小车开始,经过t=1.5s小物块通过的位移大小为多少?

五、选考3—4(4*4=16分;2*8=16分)

选对1个得2分,选对2个得3分,选对3个得4分.每选错1个扣3分,最低得分为0分 16.如图所示,一条红色光线和一条紫色光线,以不同的角度同时沿不同的

半径方向射入同一块横截面为半圆形的玻璃柱体,其透射光线都是由圆心O

点沿OC方向射出,则可知()

A.BO光线是紫光,AO光线是红光

B.BO光线是红光,AO光线是紫光

C.AO光线比BO光线穿过玻璃柱体所需时间短

D.将AO光线顺时针转动到BO光线与其重合,其透射光线仍然可能只沿OC方向射出

E.在双缝干涉实验中,若仅将入射光由AO光线变为BO光线,则干涉亮条纹间距变大

17.一列简谐横波向右传播,在其传播路径上每隔L=0.1m选取一个质点,如图甲所示,t=0时刻波恰传到质点1,并立即开始向上振动,经过时间△t=0.3s,所选取的1-9号质点间第一次出现如图乙所示的波形,则下列判断正确的是()

A.t=0.3s时刻,质点1向上运动

B.t=0.3s时刻,质点8向下运动

C.t=0至t=0.3s内,质点5运动的时间只有0.2s

D.该波的周期为0.2s,波速为4m/s

E.该波的周期为0.3s,波速为2.67m/s

18.下列说法中正确的是 ( )

A.“闻其声不见其人”是波的衍射现象

B.在太阳光照射下,水面上油膜出现彩色花纹是光的干涉现象

C.做受迫振动的物体的振动频率总是等于物体的固有频率

D.为使拍摄的水面下的景物更清晰,可利用偏振现象在照相机镜头前加一偏振片,减少反射光的影响

E.在“用单摆测定重力加速度”的实验中,当摆球摆到最高点时,按下抄表开始计时,测出30次全振动的时间,进而求出周期

19.在以下各种说法中,正确的是()

A.变化的电场一定产生变化的磁场;变化的磁场一定产生变化的电场

B.相对论认为:真空中的光速大小在不同惯性参照系中都是相同的

C.横波在传播过程中,波峰上的质点运动到相邻的波峰所用的时间为一个周期

D.机械波和电磁波本质上不相同,但它们都能发生反射、折射、干涉和衍射现象

E.如果测量到来自遥远星系上某些元素发出的光波波长比地球上这些元素静止时发光的波长长,这说明该星系正在远离我们而去

20.一列简谐横波的波形如图所示,实线表示t1=0时

刻的波形图,虚线表示t2=0.1s时刻的波形图.该波

的周期为T.

①若2T>t2-t1>T,求:该列波的传播速度.

②若波速为700m/s,求波的传播方向?

21.如图所示,玻璃棱镜ABCD可以看成是由ADE、ABE、BCD三个直角三棱镜组成.一束频率为5.3×1014H z的单色细光束从AD面入射,在棱镜中的折射光线如图中ab所示,ab与AD面的夹角α=60°.己知光在真空中的速度c=3×108m/s,玻

璃的折射率n=1.5,求:

(1)这束入射光线的入射角多大?

(2)光在棱镜中的波长是多大?(结果保留三位有效数字)

(3)该束光线第一次从CD面出射时的折射角.(结果可用反

三角函数表示)

答案

一、单选题(共5小题,共15分)

11、(1)上述步骤中:

①有重要遗漏的步骤序号是 __C____ 和 ___E___ ;

②遗漏的内容是:C中未记下两条细绳的方向,E中未说明把橡皮条的结点拉到位(2)①BC ②B;③BDE

12、①;②BC

四、计算题(本题共3小题,共8+8+10=26分)

13、解:(1)长途客车运动的速度v=20m/s,在反应时间内做匀速运动,运动的位移为:x1=vt=20×0.5m=10m

(3)当客车速度减为与狗的速度相同时,

所需时间=3.2s

司机从看到狗到速度减为与狗速度相同时,通过位移X==

=48.4m

狗通过的位移x=v(t1+t)=4×(0.5+3.2)=14.8m

X-x=48.4-14.8=33.6m>33m故狗会被撞.

答:(1)长途客车从司机发现狗至停止运动的这段时间内前进的距离为50m;

(2)狗不能摆脱被撞的噩运.

14、解:(1)分析P点受力,由平衡条件可得:

F B=G1tan37°=6 N

分析G2的受力情况由物体的平衡条件可得:

F f=G2sin37°+F B′cos37°

可求得:F f=64.8 N

(2)再分析G2的受力情况由物体的平衡条件可得:

F N+F B′sin37°=G2cos37°

可求得:F N=76.4 N.

答:木块与斜面间的摩擦力大小64.8 N;

斜面对木块的弹力大小76.4 N.

15、解:(1)物块的加速度为:a m==μg=0.2×10=2m/s2,

小车的加速度:a M===0.5m/s2.

(2)速度相等时,由a m t=v0+a M t得:

(3)在开始1s内小物块的位移:s1=.

最大速度:v=a m t=2×1m/s=2m/s.

在接下来的0.5s物块与小车相对静止,一起做加速运动

且加速度:a===0.8m/s2.

这0.5s内的位移:s2=vt+==1.1m.

经过t=1.5 s小物块通过的总位移:s=s1+s2=2.1m.

答:(1)小物块放后小物块及小车的加速度分别为2m/s2,0.5m/s2.

(2)经1s时间两者达到相同的速度.

(3)从小物块放上小车开始,经过t=1.5s小物块通过的位移大小为2.1m.

五、选考3—4(4*4=16分;2*8=16分)

选对1个得2分,选对2个得3分,选对3个得4分.每选错1个扣3分,最低得分为0分

20、解:①由图知波长λ=8m

若2T>t2-t1>T,波传播的距离在这个范围:λ<△x<2λ

若波沿x轴正方向传播,则波传播的距离为△x=λ+=8m+2m=10m

波速为v1===100m/s

若波沿x轴负向传播,则波传播的距离为△x=λ+λ=8m+6m=14m

波速为:v2==140m/s.

②在△t=0.1s内波传播的距离:△x=v△t=700×0.1m=70m

则:△x=λ=8λ+λ

所以由波形平移法可知,波沿x轴负向传播.

答:

①若波沿x轴正向传播,传播速度为100m/s,若波沿x轴负向传播为140m/s.

②波沿x轴负向传播.

21、解:(1)设光在AD面的入射角、折射角分别为i、r,则由几何知识得:r=30°

得:sini=nsinr=1.5sin30°=0.75

则得:i=arcsin0.75

(2)根据

得:

根据v=λf,得:==3.77×10-7m

(3)光路如图所示.由几何知识得ab光线在AB面的入射角为45°.

设玻璃的临界角为C,则有:

∵sin 45°>0.67,∴C<45°.

因此光线ab在AB面会发生全反射.

光线在CD面的入射角r'=r=30°,根据,得光线CD面的出射光线与法线的夹角

i'=i=arcsin0.75.

答:(1)这束入射光线的入射角arcsin0.75.

(2)光在棱镜中的波长是3.77×10-7m.

(3)该束光线第一次从CD面出射时的折射角是arcsin0.75.

浙江省宁波市镇海中学2019届高三下学期开学考试数学试题(无答案)

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