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，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，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。············································（1分）