[题目]如图所示.A.B两球质量相等.光滑斜面的倾角为.图甲中.A.B两球用轻弹簧相连.图乙中A.B两球用轻质杆相连.系统静止时.挡板C与斜面垂直.弹簧.轻杆均与斜面平行.则在突然撤去挡板的瞬间有( )A．两图中两球的加速度均为B．两图中A球的加速度均为零C．图乙中轻杆的作用力一定不为零D．图甲中B球的加速度是图乙中B球加速度的2倍题目和参考答案——青夏教育精英家教网—

【题目】如图所示，A、B两球质量相等，光滑斜面的倾角为，图甲中，A、B两球用轻弹簧相连，图乙中A、B两球用轻质杆相连，系统静止时，挡板C与斜面垂直，弹簧、轻杆均与斜面平行，则在突然撤去挡板的瞬间有（）

A．两图中两球的加速度均为

B．两图中A球的加速度均为零

C．图乙中轻杆的作用力一定不为零

D．图甲中B球的加速度是图乙中B球加速度的2倍

【答案】D

【解析】当撤去挡板时，甲图中由于弹簧的压缩量不能突变，A球还会受到不变的弹力作用，能继续平衡；而B球沿斜面方向受到重力分力mgsin和弹簧弹力mgsin两个力，所以B球会产生2gsin加速度。乙图中轻杆的形变很小，撤掉挡板后，轻杆的弹力可以发生突变。将A、B看成整体可知，A、B会沿斜面以gsin加速度下滑，可知轻杆的弹力瞬间消失。综上所述，选项D正确。

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