题目内容
如图所示,在光滑的水平桌面内有一直角坐标系xOy,在y轴正半轴与边界直线MN间有一垂直于纸面向外磁感应强度为B的匀强磁场,直线MN平行于y轴,N点在x轴上,在磁场中放置一固定在短绝缘板,其上表面所在的直线过原点O且与x轴正方向成α=30°角,在y轴上的S点左侧正前方处,有一左端固定的绝缘轻质弹簧,弹簧的右端与一个质量为m,带电量为q的带电小球接触(但不栓接),弹簧处于压缩锁定状态,在某时刻解除锁定,带电小球将垂直于y轴从S点射入磁场,垂直打在绝缘板上,并以原速率反向弹回,然后经过直线MN上的P点并垂直于MN向右离开磁场,在x轴上有一点Q,已知NP=4L,NQ=3L,则:
(1)小球带何种电荷?小球从S进入磁场后经多长时间打在绝缘板上?
(2)弹簧解除锁定前的弹性势能是多少?
(3)如果在直线MN的右侧加一方向与桌面平行的匀强电场,小球在电场力的作用下最后在Q点垂直击中x轴,那么,该匀强电场的电场强度是多少?方向如何?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552118355073.jpg)
(1)小球带何种电荷?小球从S进入磁场后经多长时间打在绝缘板上?
(2)弹簧解除锁定前的弹性势能是多少?
(3)如果在直线MN的右侧加一方向与桌面平行的匀强电场,小球在电场力的作用下最后在Q点垂直击中x轴,那么,该匀强电场的电场强度是多少?方向如何?
(1)正电(2)
(3)
,方向为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552121001033.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155211866910.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212022889.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552121001033.png)
试题分析:(1)小球只有沿Y轴负方向偏转才能与挡板相碰,根据左手则,可知小球带正电 …………………(1分)
小球进入S后做匀速圆周运动,由向心力公式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552122564604.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552122561041.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212381706.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212490644.png)
由以上三式可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212599770.png)
(2)设小球进入S后做匀速圆周运动的速度大小为V0,由向心力公式得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212661859.png)
小球进入磁场与离开磁场的速度方向都是与X轴平行向右,在磁场中轨迹如图示,
运动轨迹圆半径:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212739738.png)
在弹簧推动小球的过程中,由机械能守恒得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212817789.png)
联立求解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155211866910.png)
(3)当小球在电场力作用下沿X轴正方向走完NQ,该方向上的速度减为0时,设经过的时间为t,加速度为a1,由题意得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213020674.png)
加速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213098574.png)
该方向上的电场力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213161540.png)
同时,当小球在Y轴上向下走完PN时,设该方向上的加速度为a2,
由题意得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213223701.png)
该方向上的电场力
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213363574.png)
故小球在电场中受到电场力的大小为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213488603.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552136291460.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213691720.png)
电场强度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213753561.png)
代入数据得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155213925169.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155212022889.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241552121001033.png)
点评:做此类型题目需要,首先画出粒子在磁场中的运动轨迹,然后找出圆心,根据几何知识求半径,在运用牛顿定律解题
![](http://thumb2018.1010pic.com/images/loading.gif)
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