Question

（20分）如圖所示，物塊A的質量為M，物塊B、C的質量都是m，都可以看作質點，且m＜M＜2m。A與B、B與C用不可身長的輕線通過輕滑輪相連，A與地面用勁度系數(shù)為k的輕彈簧連接，物塊B與物塊C的距離和物塊C到地面的距離相等，假設C物塊落地后不反彈。若物塊A距滑輪足夠遠，且不計一切阻力。則：

（1）若將B與C間的輕線剪斷，求A下降多大距離時速度最大；

（2）若B與C間的輕線不剪斷，將物塊A下方的輕彈簧剪斷后，要使物塊B不與物塊C相碰，則M與m應滿足什么關系？（不計物塊B、C的厚度）

Answer 1

（1）A下降的距離為x＝x₁＋x₂＝時速度最大

（2）當M＞m時，B物塊將不會與C相碰。

解析:

（1）因為m＜M＜2m，所以開始時彈簧處于伸長狀態(tài)，其伸長量x₁，則

(2m－M)g＝kx₁····················································································································· （2分）

得x₁＝g······················································································································ （1分）

若將B與C間的輕線剪斷，A將下降B將上升，當它們的加速度為零時A的速度最大，此時彈簧處于壓縮狀態(tài)，其壓縮量x₂，則

(M－m)g＝kx₂······················································································································· （2分）

得x₂＝g······················································································································· （1分）

所以，A下降的距離為x＝x₁＋x₂＝時速度最大··································································· （2分）

（2）A、B、C三物塊組成的系統(tǒng)機械能守恒，設B與C、C與地面的距離均為L，A上升L時，A的速度達到最大，設為v，則

2mgL－MgL＝(M＋2m)v² ·································································································· （4分）

當C著地后，A、B兩物塊系統(tǒng)機械能守恒。

若B恰能與C相碰，即B物塊再下降L時速度為零，此時A物塊速度也為零，則

MgL－mgL＝(M＋m)v²········································································································ （4分）

解得：M＝m···················································································································· （3分）

由題意可知，當M＞m時，B物塊將不會與C相碰�！ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぃ�1分）