题目内容
如图所示,两根平行金属导轨MN,PQ相距为d =1.0m,导轨平面与水平面夹角为a=300导轨上端跨接一定值电阻R=1.6Ω,导轨电阻不计。整个装置处于方向垂直导轨平面向上、磁感应强度大小B=1T的匀强磁场中。金属棒ef垂直于MN、PQ静止放置,且与导轨保持良好接触,其长刚好也为d、质量m=0.1kg、电阻r=0.4Ω,距导轨底端s1=3.75m另一根与金属棒平行放置的绝缘棒gh长度也为d,质量为
,从轨道最低点以速度v0=10m/s沿轨道上滑并与金属棒发生正碰(碰撞时间极短),碰后金属棒沿导轨上滑s2=0.2m后再次静止,测得此过程中电阻R上产生的电热为Q=0.2J己知两棒与导轨间的动摩擦因数均为.
,g取10m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241237039983361.jpg)
(1) 碰后瞬问两棒的速度;
(2) 碰后瞬间金属棒的加速度;
(3) 金属棒在导轨上运动的时间。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123703967550.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123703983673.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241237039983361.jpg)
(1) 碰后瞬问两棒的速度;
(2) 碰后瞬间金属棒的加速度;
(3) 金属棒在导轨上运动的时间。
解:(1)设金属棒碰后的速度为
,
为金属棒克服安培力做的功
对金属棒碰后的过程,由动能定理:
① (2分)
再由功能关系,
② (2分)
得
(1分)
设绝缘棒与金属棒碰前的速度为
,对绝缘棒在导轨上滑动过程,由动能定理
③ (2分)
得
(1分)
设绝缘棒碰后的速度为
,选沿导轨向上的方向为正方向,
两棒碰撞过程,由动量守恒定律:
④ (2分)
得
m/s,负号表示方向沿导轨向下 (1分)
(2) 碰后金属棒切割磁感线产生的感应电动势为
,
⑤ (1分)
回路中的感应电流
⑥ (1分)
安培力的大小
⑦ (1分)
设金属棒的加速度为a,对金属棒,由牛顿第二定律:
⑧ (1分)
联立⑤⑥⑦⑧解出:
m/s2,方向沿导轨平面向下 (1分)
(3) 设金属棒在导轨上运动时间为t,在此运动过程中,安培力的冲量大小为
,沿导轨方向由动量定理:
⑨ (2分)
⑩ (1分)
由闭合电路欧姆定律:
(1分)
由法拉第电磁感应定律:
(1分)
联立⑨⑩
解得:
s (1分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704061185.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704092348.gif)
对金属棒碰后的过程,由动能定理:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241237041081273.gif)
再由功能关系,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704123574.gif)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704326400.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412370435765.gif)
设绝缘棒与金属棒碰前的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704373200.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241237043881220.gif)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704404428.gif)
设绝缘棒碰后的速度为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704420205.gif)
两棒碰撞过程,由动量守恒定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704435599.gif)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704466250.gif)
(2) 碰后金属棒切割磁感线产生的感应电动势为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704466204.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704482412.gif)
回路中的感应电流
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704513457.gif)
安培力的大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704529444.gif)
设金属棒的加速度为a,对金属棒,由牛顿第二定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704544819.gif)
联立⑤⑥⑦⑧解出:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704560257.gif)
(3) 设金属棒在导轨上运动时间为t,在此运动过程中,安培力的冲量大小为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704576217.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704576870.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704591459.gif)
由闭合电路欧姆定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704607467.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704732924.jpg)
由法拉第电磁感应定律:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704919637.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704934932.jpg)
联立⑨⑩
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241237049971061.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123704997267.gif)
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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