Question

（20分）如圖所示，物塊A的質(zhì)量為M，物塊B、C的質(zhì)量都是m，都可以看作質(zhì)點(diǎn)，且m＜M＜2m。A與B、B與C用不可身長(zhǎng)的輕線(xiàn)通過(guò)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(shè)C物塊落地后不反彈。若物塊A距滑輪足夠遠(yuǎn)，且不計(jì)一切阻力。則：

（1）若將B與C間的輕線(xiàn)剪斷，求A下降多大距離時(shí)速度最大；

（2）若B與C間的輕線(xiàn)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(yīng)滿(mǎn)足什么關(guān)系？（不計(jì)物塊B、C的厚度）

Answer 1

（1）A下降的距離為x＝x₁＋x₂＝時(shí)速度最大

（2）當(dāng)M＞m時(shí)，B物塊將不會(huì)與C相碰。

解析:

（1）因?yàn)?i>m＜M＜2m，所以開(kāi)始時(shí)彈簧處于伸長(zhǎng)狀態(tài)，其伸長(zhǎng)量x₁，則

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

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

若將B與C間的輕線(xiàn)剪斷，A將下降B將上升，當(dāng)它們的加速度為零時(shí)A的速度最大，此時(shí)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(shí)速度最大··································································· （2分）

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

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

當(dāng)C著地后，A、B兩物塊系統(tǒng)機(jī)械能守恒。

若B恰能與C相碰，即B物塊再下降L時(shí)速度為零，此時(shí)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(dāng)M＞m時(shí)，B物塊將不會(huì)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）