题目内容
(16分)如图所示,竖直平面内的光滑弧形轨道的底端恰好与光滑水平面相切。质量为M=2.0kg的小物块B静止在水平面上。质量为
=1.0kg的小物块A从距离水平面高
=0.45m的P点沿轨道从静止开始下滑,经过弧形轨道的最低点Q滑上水平面与B相碰,碰后两个物体以共同速度运动。取重力加速度
=10m/s2。求
(1)A经过Q点时速度的大小
;
(2)A与B碰后速度的大小
;
(3)碰撞过程中系统(A、B)损失的机械能
。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241210402522685.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040174204.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040190200.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040190197.gif)
(1)A经过Q点时速度的大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040206204.gif)
(2)A与B碰后速度的大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040221185.gif)
(3)碰撞过程中系统(A、B)损失的机械能
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040237333.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241210402522685.jpg)
(1)3.0m/s
(2)1.0m/s
(3)3.0J
(2)1.0m/s
(3)3.0J
(1)A从P滑到Q的过程中,根据机械能守恒定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040284574.gif)
解得A经过Q点时速度的大小
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040284634.gif)
(2)A与B相碰,根据动量守恒定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040299559.gif)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040315700.gif)
(3)根据能量守恒定律得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040330818.gif)
解得A与B碰撞过程中系统损失的机械能
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824121040346427.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目