（1）因为m＜M＜2m，所以开始时弹簧处于伸长状态，其伸长量x1，则
(2m－M)g＝kx1····················································································································· （2分）
得x1＝g······················································································································ （1分）
若将B与C间的轻线剪断，A将下降B将上升，当它们的加速度为零时A的速度最大，此时弹簧处于压缩状态，其压缩量x2，则
(M－m)g＝kx2······················································································································· （2分）
得x2＝g······················································································································· （1分）
所以，A下降的距离为x＝x1＋x2＝时速度最大··································································· （2分）
（2）A、B、C三物块组成的系统机械能守恒，设B与C、C与地面的距离均为L，A上升L时，A的速度达到最大，设为v，则
2mgL－MgL＝(M＋2m)v2 ·································································································· （4分）
当C着地后，A、B两物块系统机械能守恒。
若B恰能与C相碰，即B物块再下降L时速度为零，此时A物块速度也为零，则
MgL－mgL＝(M＋m)v2········································································································ （4分）
解得：M＝m···················································································································· （3分）
由题意可知，当M＞m时，B物块将不会与C相碰。············································（1分） | 
