题目内容
(21分)在光滑绝缘的水平面上建有如图所示的平面直角坐标系,在此水平面上可视为质点的不带电小球a静止于坐标系的原点O,可视为质点的带正电小球b静止在坐标为(0,﹣h)的位置上。现加一方向沿y轴正方向、电场强度大小为E、范围足够大的匀强电场,同时给a球以某一速度使其沿x轴正方向运动。当b球到达坐标系原点O时速度为v0,此时立即撤去电场而改加一方向垂直于绝缘水平面向上、磁感应强度大小为B、范围足够大的匀强磁场,最终b球能与a球相遇。求:
(1)b球的比荷q/m;
(2)从b球到达原点O开始至b球与a球相遇所需的时间;
(3)b球从开始位置运动到原点O时,a球的位置。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214365153491.jpg)
(1)b球的比荷q/m;
(2)从b球到达原点O开始至b球与a球相遇所需的时间;
(3)b球从开始位置运动到原点O时,a球的位置。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214365153491.jpg)
(1)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436530552.gif)
(2)t=
……
(3)(
,0) (
……)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436530552.gif)
(2)t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436562752.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436577307.gif)
(3)(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436593727.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436577307.gif)
(1)b球受电场力作用做匀加速运动,由动能定理得:
(3分)则b球的比荷为
(2分)
(2)b球运动到原点后将在水平面上做匀速圆周运动
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214366714588.jpg)
所以有:
(2分)
(2分)
b球只能与a球相遇在图中的S处,相遇所需时间为
(2分)
t=
…… (2分)
(3)a球开始运动到与b球相遇所用时间
(1分)其中
(1分)
a球通过的路程为OS=2R (1分)所以可得a球的速度:v=
(1分)
故v=
(1分)
则a球在x轴上的坐标为
(
……)(2分)
a球的位置为(
,0) (
……)(1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436640574.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436530552.gif)
(2)b球运动到原点后将在水平面上做匀速圆周运动
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241214366714588.jpg)
所以有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436686571.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436702740.gif)
b球只能与a球相遇在图中的S处,相遇所需时间为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436718575.gif)
t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436562752.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436577307.gif)
(3)a球开始运动到与b球相遇所用时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436764272.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436796442.gif)
a球通过的路程为OS=2R (1分)所以可得a球的速度:v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436811290.gif)
故v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436827727.gif)
则a球在x轴上的坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436842874.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436577307.gif)
a球的位置为(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436593727.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121436577307.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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