题目内容
如图12所示质量为M的小物块A静止在离地面高h的水平桌面的边缘,质量为m的小物块B沿桌面向A运动并以速度v0与之发生正碰(碰撞时间极短)。碰后A离开桌面,其落点离出发点的水平距离为L。碰后B反向运动。已知B与桌面的动摩擦因数为μ,重力加速度为g,桌面足够长。求
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413515646310019.png)
(1)碰后A、B小物块分离的瞬间速率各是多少?
(2)碰后小物块B后退的最大距离是多少?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082413515646310019.png)
(1)碰后A、B小物块分离的瞬间速率各是多少?
(2)碰后小物块B后退的最大距离是多少?
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241351567591453.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241351567131083.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241351567591453.png)
本题考查碰撞前后的动量守恒问题,先根据平抛运动求出碰撞后的速度,再由动量守恒列式求解
(1)设:A、B碰后的瞬时速度大小分别是V和v,由动量守恒定律有:
……………………………………………………………………①
对于A的平抛运动,有:
L=V·T………………………………………………………………………………②
…………………………………………………………………………③
解得:
,………………………………………………………………④
……………………………………………………………………⑤
(2)设B球后退最大距离为S,则由动能定理,有
……………………………………………………………………⑥
…………………………………………………………⑦
(1)设:A、B碰后的瞬时速度大小分别是V和v,由动量守恒定律有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135156791742.png)
对于A的平抛运动,有:
L=V·T………………………………………………………………………………②
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135156837678.png)
解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135156900741.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241351567131083.png)
(2)设B球后退最大距离为S,则由动能定理,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824135157227852.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241351567591453.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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