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一个平板小车置于光滑水平面上，其右端恰好和一个

1
4
光滑圆弧固定轨道AB的底端等高对接，如图所示．已知小车质量M=3.0kg，长L=2.06m，圆弧轨道半径R=0.8m．现将一质量m=1.0kg的小滑块，由轨道顶端A点无初速释放，滑块滑到B端后冲上小车．滑块与小车上表面间的动摩擦因数μ=0.3  （取g=10m/s2）

试求：

（1）滑块到达底端B时，对轨道的压力大小．

（2）通过计算判断滑块是否能滑离小车？若能滑离，求滑离时小车的速度；若不能滑离，求最终滑块相对小车的滑行距离．

答案

（1）滑块从A端下滑到B端，由机械能守恒得

    mgR=

1
2
m
v20

得v0=

2gR
=4m/s

在B点，由牛顿第二定律得FN-mg=m

v20
R
    

解得轨道对滑块的支持力FN=3 mg=30 N   

由牛顿第三定律可知，滑块对轨道的压力为30 N 

（2）滑块滑上小车后，假设滑块没有滑出小车二者同速，设速度为v，

由动量守恒：mv0=（M+m）v，得v=1m/s     

由能的转化和守恒得：μmg•△s=

1
2
m
v20
-
1
2
(M+m)v2

滑块在小车上滑行长度△s=2m＜L=2.06m

即滑块不能滑离小车

答：（1）滑块到达底端B时，对轨道的压力大小是30N．

（2）滑块不能滑离小车，最终滑块相对小车的滑行距离是2m．

单项选择题

下列运算符不能在C++中重载的是( )。

A．：

B．-

C．＜=

D．+
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