题目内容
(20分)如图,O、A、B为同一竖直平面内的三个点,OB沿竖直方向,
,
.将一质量为m的小球以一定的初动能自O点水平向右抛出,小球在运动过程中恰好通过A点。使此小球带电,电荷量为q(q>0),同时加一匀强电场,场强方向与
所在平面平行,现从O点以同样的初动能沿某一方向抛出此带点小球,该小球通过了A点,到达A点时的动能是初动能的3倍;若该小球从O点以同样的初动能沿另一方向抛出,恰好通过B点,且到达B点的动能为初动能的6倍,重力加速度大小为g。求
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049236912030.png)
(1)无电场时,小球达到A点时的动能与初动能的比值;
(2)电场强度的大小和方向。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923628664.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923660705.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923660517.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049236912030.png)
(1)无电场时,小球达到A点时的动能与初动能的比值;
(2)电场强度的大小和方向。
(1)
(2) ![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923784837.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923769710.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923784837.png)
试题分析:(1)无电场时,小球运动为平抛运动,设初速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923800323.png)
水平方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923831801.png)
竖直方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923847968.png)
解得时间
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923862760.png)
竖直方向速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923894869.png)
经过A点的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923909994.png)
初动能
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923925807.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049239401095.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923769710.png)
(2)从O点到A点,没有加电场时,根据动能定理有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049239721568.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923987931.png)
加电场后通过A点根据动能定理则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049240031825.png)
联立可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924018876.png)
加电场后恰好通过B点,根据动能定理则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049240343981.png)
联立可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924050747.png)
整理可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924065696.png)
设直线OB上面的M点与A点等电势,如下图
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049240819639.png)
根据匀强电场中沿任意一条直线电势都均匀变化,所以有
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049241121038.png)
根据MA电势相等该条直线即为匀强电场的等势面,那么垂线OC与MA垂直即为电场线,设电场线与OB的夹角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924128310.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924143480.png)
根据正电荷从O到B电场力做正功,判断电场方向与竖直方向成
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004924143480.png)
根据从O到A电场力做功
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250049241742276.png)
整理可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825004923784837.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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