题目内容
倾角θ为37°的斜面与水平面如图平滑相接,A、B两完全相同的物块静置于斜面上相距
,B距斜面底端的P点的距离
,物块与斜面及水平面的动摩擦因数均为
.现由静止释放物块A后1秒钟再释放物块B.设AB碰撞的时间极短,碰后就粘连在一起运动.试求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912345407.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912361403.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912377267.gif)
(1)B物块释放后多长时间,AB两物块发生碰撞?
(2)AB最后停在距斜面底端P点多远处?取
,
,
.![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241149124231204.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912392450.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912408447.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912408458.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241149124231204.gif)
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912455416.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912439248.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912455416.gif)
(1)设B物体的质量为m,加速下滑时的加速度为a,其在斜面上时的受力情况如图所示。由牛顿第二定律得:
解得
设B物块释放后经t秒A追上B与其在斜面上相碰,由两者的位移关系得:
代入数据得:
在此1.5s内,B物体下滑的位移
因为
,AB确实是在斜面上发生碰撞。
(2)A碰前的速度
B碰前的速度
由于碰撞时间极短,两者碰撞近似动量守恒,设碰后两者的共同速度为v,则
代入数据得![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912704525.gif)
解得
AB相碰时距斜面底端的高度
设AB滑下斜面后停在距P点S3远处。由动能定理得:
代入数据解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912470573.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912486538.gif)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912501791.gif)
设B物块释放后经t秒A追上B与其在斜面上相碰,由两者的位移关系得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912517646.gif)
代入数据得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912439248.gif)
在此1.5s内,B物体下滑的位移
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912611814.gif)
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912642369.gif)
(2)A碰前的速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912657750.gif)
B碰前的速度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912673611.gif)
由于碰撞时间极短,两者碰撞近似动量守恒,设碰后两者的共同速度为v,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912689526.gif)
代入数据得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912704525.gif)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912720384.gif)
AB相碰时距斜面底端的高度
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912735925.gif)
设AB滑下斜面后停在距P点S3远处。由动能定理得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241149127511400.gif)
代入数据解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114912455416.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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