18.(1)h≥0.5m(2)45N(3)10【解析】本题考动量守恒定律和能量守恒定律的应用。(1)小球恰能完成一次完整的圆周运动时,设它到最高点的速度为vo,在最高点,仅有重力充当向心力,则有mg=m(1分)在小球从最低点运动到最高点的过程中,机械能守恒,以水平面为零势能面,并设小球在最低点的速度为n,则有m=mg·(1分解得t1=√5m/s(1分设滑块从ho高处运动到将与小球碰撞时速度为n2,对滑块由能星守恒定律有(1分因滑块与小球发生弹性碰撞,故碰潼后速度交换,则n2=√5m′s(1分解得h=0.5m(1分因此滑块B从斜面下滑的高度h≥0.5m(1分)(2)若滑块从h=5m处下滑到将要与小球碰撞时速度为u,同理有mgh′解得u=√95Im s(1分)滑块与小球碰后的瞬间,同理滑块静止,小球以u′=√95m/s的速度开始做圆周运动,因为a>v1,则小球能运动到圆周运动的最高点,根据小球从最低点运动到最高点的过程中,机械能守恒以水平面为零势能面,设小球在最高点的速度为v3,则有o mu=mg 2L+2 mu3(1分}设小球运动到最高点时绳的拉力大小为T,重力与绳的拉力的合力充当向心力,则有T十L(1分)T+mg=m(1分解得T=45N(1分)(3)滑块和小球最后一次碰撞时速度为t=5m/s滑块最后停在水平面上,设它通过的路程为s,同理有mgh=m+mgs(1分小球做完整圆运动的次数为n=(1分解得s′=19m,n=10

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