题目内容
(8分)在光滑水平地面上有两个相同的弹性小球A、B,质量都为m.现B球静止,A球向B球运动,发生正碰.已知碰撞过程中总机械能守恒,两球压缩最紧时的弹性势能为Ep,求:碰前A球的速度大小等于多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241422288421218.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241422288421218.png)
v0=2
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142228920617.png)
设碰前A球的速度v0,两个弹性小球发生正碰,当二者共速时,弹簧弹性势能最大,由动量守恒得mv0=2mv,Ep=
mv
-
×2mv2,解得v0=2
,本题考查碰撞过程中的动量守恒定律的应用,当两小球速度相同时速度相同,根据动量守恒和能量守恒列式求解
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142229045338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142229248293.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142229045338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824142228920617.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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