高考模拟试题及答案

2011年高考模拟试题及参考答案

命题人：卢锦亮

一、单项选择题：

1、科学家发现在月球上含有丰富的3

2He （氦3），它是一种高效、清洁、安全的核聚变燃料，其参与的一种核聚变反应的方程式为3

2He +3

2He →1

1

2H +42He ，关于3

2He 聚变下列

表述正确的是A.聚变反应不会释放能量 B.聚变反应产生了新的原子核 C ．聚变反应没有质量亏损 D.目前核电站都采用3

2He 聚变反应发电 2、下列关于热学现象和热学规律说法正确的是：

A ．根据热力学第二定律可知热量能够从高温物体传到低温物体，但不可能从低温物体传到高温物体

B ．用活塞压缩气缸里的空气，对空气做功2.0×105

J ，同时空气的内能增加了1.5×105

J ，

则空气从外界吸收热量0.5×105

J

C ．物体的温度为0℃时，物体分子的平均动能为零

D ．一定质量的气体，保持温度不变，压强随体积增大而减小的微观原因是：单位体积

内的分子数减小

3．质量为M 的人造地球卫星在距地球表面高度为R 的轨道上绕地球做匀速圆周运动，其中

R 为地球半径，设地球表面的重力加速度为g ，以下说法正确的是 A ．卫星的线速度为2

gR

v =

B ．一个质量m 的物体在该卫星内所受的重力为0

C ．如果该卫星经变轨到距地面高为2R 的新轨道上运动则其运动周期变小

D ．该卫星是地球的同步卫星

4．两根光滑的金属导轨，平行放置在倾角为θ的斜面上，导轨的左端接有电阻R ，导轨的电阻可忽略不计。斜面处在匀强磁场中，磁场方向垂直于斜面向上。质量为m ，电阻可不计的金属棒ab 在沿着斜面与金属棒垂直的拉力F 作用下沿导轨匀速上滑，并上升h 高度，如图所示，在这个过程中 A ．作用在金属棒上各个力的合力所做的功等于电阻R 上产生的

焦耳热 B ．作用在金属棒上各个力的合力所做的功等于mgh 与电阻R 上

产生的焦耳热之和

C ．恒力F 与重力的合力所做的功等于电阻R 上产生的焦耳热

D ．恒力F 与安培力的合力所做的功等于零

h F

b a θB

二、双项选择题

5．某物体受到合力F的作用，由静止开始运动，其v-t图

象如图所示，由图可知：

A．该物体只向一个方向运动

B．该物体所受的合力F方向不断变化，而大小始终不变

C．3 s末物体回到原出发点

D．2 s～4 s内，物体的动量变化量不等于零，合力F

对物体所做的功等于零

6．如图所示，平行金属板中的带电质点P处于静止状态，不考虑电流表和电压表对电路的影响，当滑动变阻器R4滑片向b端移动时

（）

A．电流表读数增大

B．电压表读数增大

C．R3上消耗的电功率增大

D．质点P将向下运动

7、在如图甲所示的电路中，电阻R的阻值为50Ω，在ab间加上图乙所示的正弦交流电，

则下面说法中正确的是：

A、电流表示数为2A

B、产生该交流电的线圈在磁场中转动的角速度为3．14rad/s

C、如果产生该交流电的线圈转速提高一倍，则电流表的示数也增大一倍

D、1秒内通过R的电流方向变化50次。

8、如图中虚线表示匀强电场的等势面1、2、3、4。一带正电的粒子只在电场力的作用下从电场中的a点运动到b点，轨迹如图中实线所示。由此可判断

A．4等势面电势最高

B．粒子从a运动到b过程中动能增大

C．粒子从a运动到b过程中电势能增大

D．在运动中粒子的电势能与动能之和不变

9．地铁是大都市重要的交通工具之一,在地铁列车上通常有提示列车加速状态的指示灯，旅客可以通过指示灯把握好自己的坐姿,防止自己从座位上摔出去,其结构如图所示,M是质量较大的金属块，左右两端分别与金属丝制作的弹簧相连，并套在光

滑水平金属杆上，当M平衡时两根弹簧均处于原长，此时两灯都不亮。

下列说法正确的是

A、当地铁列车加速前进时，甲灯可能变亮

B、当乙灯突然变亮时, 旅客有向车后倒的趋势

M

V

头

车

后

甲灯

乙灯

C、当乙灯亮时, 旅客受的合力方向向车后

D、当甲灯亮时，列车可能是减速进站了

三、实验题：

10、有一同学测量小车从斜面上下滑所受到的阻力大小，他利用一打点计时器固定在斜面上某处，一小车拖着穿过打点计时器的纸带从斜面上滑下，如图8所示．图9是打出的纸带的一段,已量出各相邻计数段的长度分别为：S1、S2、S3、S4、S5、S6、S7、S8 .

(1)已知打点计时器使用的交流电频率f =50Hz，打点的周期为 s. 用以上已知的物理量表示小车下滑的加速度算式为a = .

（2）为了求出小车在下滑过程中所受的阻力，还需测量的物理量有 . （3）用加速度a及其他需要测得的量表示阻力的计算式为f = .

11、在做“描绘小灯泡的伏安特性曲线”的实验中，采用的仪器有：电源E（电动势6V，内阻不计）；额定电压4.0V的小灯泡L；电流表○A（量程0.6A，内阻约2Ω）；电压表表○V（量程5V，内阻约15kΩ）；滑动变阻器R（最大阻值15Ω）；开关S，导线若干。实验中要求加在小灯泡上的电压从零到小灯泡的额定电压连续可调。

（I）．在做实验时，连成的实物电路图如下：

请你指出上述实验电路图中的错误

．．之处：

①；②

（II）．请你在虚线方框中画出实验所用的正确电路原理图。

S1 S2 S3 S4 S5 S6 S7 S8

图8 图9

（Ⅲ）某位同学测得小灯泡的伏安特性曲线如图4所示。图象是一条曲线而不是直线的原因

是————————————————；某次测量时，电流表指针位置如图5所示，电流表读数为 A

，此时小灯泡的实际功率为 W 。

三、计算题：

20．(16分) 一轻质细绳一端系一质量为20

1

m kg 的小球A ，另一端挂在光滑水平轴O 上，O 到小球的距离为L =0.1m ，小球跟水平面接触，但无相互作用，在球的两侧等距离处分别固定一个光滑的斜面和一个挡板，如图所示，水平距离s 为2m ，动摩擦因数为0.25．现有一小滑块B ，质量也为m ，从斜面上滑下，与小球碰撞时交换速度，与挡板碰撞不损失机械能．若不计空气阻力，并将滑块和小球都视为质点，g 取10m/s 2，试问：

（1）若滑块B 从斜面某一高度h 处滑下与小球

第一次碰撞后，使小球恰好在竖直平面内 做圆周运动，求此高度h .

（2）若滑块B 从h =5m 处滑下，求滑块B 与

小球第一次碰后瞬间绳子对小球的拉力． （3）若滑块B 从h =5m 处下滑与小球碰撞后，

小球在竖直平面内做圆周运动，求小球做完整圆周运动的次数．

13、如图，在x oy 平面第一象限内的MN 和x 轴之间有平行于y 轴的匀强电场和垂直于x oy 平面的匀强磁场，y 轴上离坐标原点4L 的A 点处有一电子枪，可以沿+x 方向射出速度为v 0的电子（质量为m ，电量为e ）不计电子的重力。.如果电场和磁场同时存在，电子将做匀速直线运动，如果撤去电场，只保留磁场，电子将从x 轴上距坐标原点3L 的C 点离开磁场.求：(1)磁感应强度B 和电场强度E 的大小和方向；

(2)如果撤去磁场，只保留电场，电子将从D 点（图中未标出）离开电场，求D 点的坐标；（3）电子通过D 点时的动能。

O s /2 s /2 A h

B

（3）从A 点到D 点，由动能定理 20212υm E L Ee kD -=? （2分）求出2

50

57υm E kD =

2011年高考模拟试题答案：

一、单项选择题：1、B 、2、D 、3、A 、4、C

二、双项选择题：5、CD 、6、AD 、7、AC 、8、BD 、9、BD 三、实验题： 10、（1）略

（2）．①电流表没有接成外接电路 ②滑动变阻器没有接成分压电路。 Ⅱ．见右图

Ⅲ．①灯丝电阻随温度而改变（灯丝电阻随I 或

随U 改变亦可）

②10Ω（9.5Ω到10.5Ω之间都给分）

11、 （16分）解：（1）小球刚能完成一次完整的圆周

运动，它到最高点的速度为v 1，在最高点，仅有重力充当向心力，则有 L

v m

mg 2

1=

①

在小球从最低点运动到最高点的过程中，机械能守恒，并设小球在最低点速度为v ，则又有

2122

1

221mv L mg mv +?= ②

解①②有s m v /5=

滑块从h 1高处运动到将与小球碰撞时速度为v ，对滑块由能的转化及守恒定律有

2

12

12mv s mg mgh +?

=μ 因碰撞后速度交换s m v /5=

，解上式有h 1=0．5m

（2）若滑块从h =5m 处下滑到将要与小球碰撞时速度为u ，同理有

2

212s mg mu mgh ?+=

μ ③ 解得s m u /95=

滑块与小球碰后的瞬间，同理滑块静止，小球以s m u /95=的速度开始作圆周运动，

绳的拉力T 和重力的合力充当向心力，则有 L

u m mg T 2

=-④ 解④式得T =48N

（3）滑块和小球最后一次碰撞时速度为s m v /5=，滑块最后停在水平面上，它通

过的路程为s '，同理有 s mg mv mgh '+=

μ2

21

⑤ 小球作完整圆周运动的次数为12+-

'=s

s s n ⑥ 解⑤、⑥得m s 19='，n =10次 12.（20分）解：（1）只有磁场时，电子运动轨迹如图1所示（1分）

洛仑兹力提供向心力R

m

Be 2

0υυ= （1分）

由几何关系 222)4()3(R L L R -+=（2分） 求出eL m B 2580

υ=

垂直纸面向里（2分） 电子做匀速直线运动0υBe Ee = （1分） 求出eL

m E 25820υ=

沿y 轴负方向（2分）

（2）只有电场时，电子从MN 上的D 点离开电场，如图2所示（1分）

设D 点横坐标为x t x 0υ= （2分） 2

212t m

eE L = （2分） 求出D 点的横坐标为L L x 5.32

2

5≈= （1分）纵坐标为L y 6= （1分）

（3）从A 点到D 点，由动能定理 20212υm E L Ee kD -

=? （2分）求出2

50

57υm E kD =

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