(2012江苏四市联考)如图所示,A、B两小球由绕过轻质定滑轮的细线

2012江苏四市联考)如图所示,AB两小球由绕过轻质定滑轮的细线相连,A放在固定的光滑斜面上,BC两小球在竖直方向上通过劲度系数为k的轻质弹簧相连,C球放在水平地面上。现用手控制住A,并使细线刚刚拉直但无拉力作用,并保证滑轮左侧细线竖直、右侧细线与斜面平行。已知A的质量为4mBC的质量均为m,重力加速度为g,细线与滑轮之间的摩擦不计,开始时整个系统处于静止状态。释放A后,A沿斜面下滑至速度最大时C恰好离开地面。下列说法正确的是

       A.斜面倾角α=30°

       BA获得最大速度为

       CC刚离开地面时,B的加速度最大

       D.从释放AC刚离开地面的过程中,AB两小球组成的系统机械能守恒

 


答案

答案:AB解析:释放A后,A沿斜面下滑至速度最大时C恰好离开地面,细绳中拉力等于4mgsinα。弹簧中弹力等于mg,由4mgsinα=mg+mg解得斜面倾角α=30°,选项A正确;释放A前,弹簧压缩x=mg/kA沿斜面下滑至速度最大时弹簧拉伸x‘=mg/k由机械能守恒定律,4mg·2xsinα-mg·2x=·4mv2+·mv2,解得A获得最大速度为v=,选项B正确;C刚离开地面时,B的加速度为零,选项C错误;从释放AC刚离开地面的过程中,AB两小球、地球、弹簧组成的系统机械能守恒,选项D错误。

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