题目内容
（20分）如图所示，物块A的质量为M，物块B、C的质量都是m，都可以看作质点，且m＜M＜2m。A与B、B与C用不可身长的轻线通过轻滑轮相连，A与地面用劲度系数为k的轻弹簧连接，物块B与物块C的距离和物块C到地面的距离相等，假设C物块落地后不反弹。若物块A距滑轮足够远，且不计一切阻力。则：
（1）若将B与C间的轻线剪断，求A下降多大距离时速度最大；
（2）若B与C间的轻线不剪断，将物块A下方的轻弹簧剪断后，要使物块B不与物块C相碰，则M与m应满足什么关系？（不计物块B、C的厚度）
（1）A下降的距离为x＝x1＋x2＝时速度最大
（2）当M＞m时，B物块将不会与C相碰。
解析:
（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）
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