![](https://wkrtcs.bdimg.com/rtcs/image?w=624&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"624","h":"931","dataType":"jpeg","c":"word/media/image1.png","imgOriW":1296,"imgOriH":1935})
参考答案
题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
答案 | D | B | C | D | B | A | AD | CD | AC | AC |
![](https://wkrtcs.bdimg.com/rtcs/image?w=189.33333333333&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"189.33333333333","h":"6.6666666666667","dataType":"gif","c":"word/media/image7.png"})
11.（6分） (1) (2分) (2) 0. 2(2分) 0.25(2分)
12.（10分）
（1） D （1分）
（2）实验原理图如图所示（3分）
（3）答案如图（2分）
（4）1.48（1.47~1.49）（2分）
0.30(0.25~0.35均正确) （2分）
13.（10分)解：（1）由图像得![](https://wkrtcs.bdimg.com/rtcs/image?w=240&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"240","h":"86","dataType":"png","c":"word/media/image10.png","imgOriW":243,"imgOriH":89})
………1分
上滑的位移![](https://wkrtcs.bdimg.com/rtcs/image?w=202&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"202","h":"86","dataType":"png","c":"word/media/image11.png","imgOriW":205,"imgOriH":89})
………1分
由牛顿第二定律知 ![](https://wkrtcs.bdimg.com/rtcs/image?w=192&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"192","h":"19","dataType":"png","c":"word/media/image12.png","imgOriW":191,"imgOriH":16})
………2分 解得 ………1分
（2）由于![](https://wkrtcs.bdimg.com/rtcs/image?w=58&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"58","h":"19","dataType":"png","c":"word/media/image14.png","imgOriW":61,"imgOriH":16})
，物块无法在斜面上静止，沿斜面下滑由牛顿第二定律知
![](https://wkrtcs.bdimg.com/rtcs/image?w=269&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"269","h":"86","dataType":"png","c":"word/media/image15.png","imgOriW":268,"imgOriH":89})
………1分
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=67&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"67","h":"29","dataType":"png","c":"word/media/image16.png","imgOriW":66,"imgOriH":30})
………1分
有![](https://wkrtcs.bdimg.com/rtcs/image?w=317&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"317","h":"86","dataType":"png","c":"word/media/image17.png","imgOriW":315,"imgOriH":89})
………1分
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=86&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"86","h":"86","dataType":"png","c":"word/media/image18.png","imgOriW":91,"imgOriH":89})
………1分
即物块在斜面上运动的时间![](https://wkrtcs.bdimg.com/rtcs/image?w=77&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"77","h":"86","dataType":"png","c":"word/media/image19.png","imgOriW":80,"imgOriH":89})
s ………1分
14．（10分）解：（1）小球沿直线AO向上运动，则合力方向与AO共线，电场力水平向右，重力竖直向下，则合力沿OA斜向下，则Eq=mgtan370，………2分
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=125&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"125","h":"77","dataType":"png","c":"word/media/image20.png","imgOriW":127,"imgOriH":78})
………1分
（2）设动能减到零时向上的位移为L，
由动能定理：![](https://wkrtcs.bdimg.com/rtcs/image?w=240&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"240","h":"86","dataType":"png","c":"word/media/image21.png","imgOriW":236,"imgOriH":89})
……2分
由功能关系：![](https://wkrtcs.bdimg.com/rtcs/image?w=413&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"413","h":"86","dataType":"png","c":"word/media/image22.png","imgOriW":409,"imgOriH":89})
………2分
联立解得Wf=![](https://wkrtcs.bdimg.com/rtcs/image?w=38&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"38","h":"86","dataType":"png","c":"word/media/image23.png","imgOriW":41,"imgOriH":89})
E0………2分
则小球回到P时，机械能损失2 Wf=E0，即动能为E0………1分
15．（12分）解：（1）A、B碰撞过程动量守恒 则有![](https://wkrtcs.bdimg.com/rtcs/image?w=269&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"269","h":"86","dataType":"png","c":"word/media/image24.png","imgOriW":266,"imgOriH":89})
……2分
其中![](https://wkrtcs.bdimg.com/rtcs/image?w=106&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"106","h":"86","dataType":"png","c":"word/media/image25.png","imgOriW":105,"imgOriH":89})
，解得……2分
（2）碰撞后C向右滑上传送带，因![](https://wkrtcs.bdimg.com/rtcs/image?w=29&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"29","h":"86","dataType":"png","c":"word/media/image27.png","imgOriW":33,"imgOriH":89})
顾C在传送带上先做匀减速运动，加速度大小
![](https://wkrtcs.bdimg.com/rtcs/image?w=106&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"106","h":"29","dataType":"png","c":"word/media/image28.png","imgOriW":110,"imgOriH":30})
……1分
设C的速度减小到跟传送带速度相等时向右运动的距离为x，
根据运动学公式![](https://wkrtcs.bdimg.com/rtcs/image?w=182&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"182","h":"86","dataType":"png","c":"word/media/image29.png","imgOriW":184,"imgOriH":89})
............1分
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=106&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"106","h":"19","dataType":"png","c":"word/media/image30.png","imgOriW":105,"imgOriH":16})
……1分 这一过程的时间……1分
C与传送带共速后匀速运动到P点的时间为![](https://wkrtcs.bdimg.com/rtcs/image?w=144&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"144","h":"86","dataType":"png","c":"word/media/image32.png","imgOriW":146,"imgOriH":89})
……1分
C进入圆弧轨道后，设上升的最大高度为h ,根据机械能守恒定律有
![](https://wkrtcs.bdimg.com/rtcs/image?w=518&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"518","h":"86","dataType":"png","c":"word/media/image33.png","imgOriW":522,"imgOriH":89})
……2分
解得.即C不会从圆弧轨道滑出。........1分
16．（12分）解：（1）电子从A到C的过程中，由动能定理得: ………1分
………1分
联立解得![](https://wkrtcs.bdimg.com/rtcs/image?w=202&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"202","h":"86","dataType":"png","c":"word/media/image34.png","imgOriW":199,"imgOriH":89})
…………1分
电子做类平抛运动，y方向匀速运动，x方向匀加速运动，
则……2分
解得![](https://wkrtcs.bdimg.com/rtcs/image?w=29&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"29","h":"19","dataType":"png","c":"word/media/image35.png","imgOriW":68,"imgOriH":39})
，则C点的坐标为………1分
（2）由洛伦兹力提供向心力有:![](https://wkrtcs.bdimg.com/rtcs/image?w=192&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"192","h":"86","dataType":"png","c":"word/media/image37.png","imgOriW":193,"imgOriH":89})
……1分 解得半径……1分
![](https://wkrtcs.bdimg.com/rtcs/image?w=175&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"175","h":"144.46666666667","dataType":"png","c":"word/media/image39.png","imgOriW":149,"imgOriH":123})
电子在磁场中偏转120°后垂直于ON射出，如图：
磁场最小半径为![](https://wkrtcs.bdimg.com/rtcs/image?w=106&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"106","h":"38","dataType":"png","c":"word/media/image40.png","imgOriW":223,"imgOriH":88})
……2分，
可得![](https://wkrtcs.bdimg.com/rtcs/image?w=67&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"67","h":"48","dataType":"png","c":"word/media/image41.png","imgOriW":145,"imgOriH":92})
……1分
圆形磁场的最小面积![](https://wkrtcs.bdimg.com/rtcs/image?w=67&md5sum=88d194af6950f5aa7c353dec1c88fc70&sign=a01a911754&rtcs_flag=1&rtcs_ver=3&l=webapp&bucketNum=122&ipr={"t":"img","w":"67","h":"29","dataType":"png","c":"word/media/image42.png","imgOriW":142,"imgOriH":57})
得……1分
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