Question

如圖所示，質(zhì)量為3m的足夠長木板 C 靜止在光滑水平面上，質(zhì)量均為 m 的兩個物體 A、B 放在 C 的左端，A、B 間相距 s₀，現(xiàn)同時對A、B施加水平向右的瞬間沖量而使之分別獲得初速度v₀和2v₀，若A、B與C之間的動摩擦因數(shù)分別為 μ 和 2 μ ，則：

（1）最終A、B、C的共同速度為多大

（2）求A達到最小速度時，系統(tǒng)產(chǎn)生的熱量Q。

Answer 1

（1）0.6v₀；（2）1.5mv₀²

解析:

（1）對A、B、C組成的系統(tǒng)，最終相對靜止，設(shè)共同速度為v_共，由動量守恒定律：

mv₀＋2mv₀＝（m＋m＋3m）v_共······································································· ①（4分）

解得：v_共＝0.6v₀ ······················································································· ②（1分）

（2）A、B相對C向右勻減速運動階段，C做勻加速運動，設(shè)A的加速度為a_A，B的加速度為a_B，C的加速度為a_C ，設(shè)t時刻A、C達到共同速度v_A，B的速度為v_B

由牛頓第二定律：

a_A＝gμ ··································································································· ③（1分）

a_B＝2gμ ································································································· ④（1分）

a_C＝＝gμ ·························································································· ⑤（1分）

t時間內(nèi)，對A，由速度公式：v_A＝v₀－a_At ························································ ⑥（1分）

對B：v_B＝2v₀－a_Bt ···················································································· ⑦（1分）

對C：v_A＝a_C t ·························································································· ⑧（1分）

解得：v_A＝0.5v₀     ·················································································· ⑨（1分）

v_B＝v₀ ···································································································· ⑩（1分）

此后A、C一起勻加速運動，B繼續(xù)以加速度a_B減速運動，故運動過程中A的最小速度為v_A＝0.5v₀

············································································································ ⑾（1分）

根據(jù)系統(tǒng)的能量守恒定律，系統(tǒng)產(chǎn)生的熱量

Q＝mv₀²＋m(2v₀)²－mv??_A²－×3mv??_A²－mv??_B² ·················································· ⑿（4分）

解得 Q＝1.5mv₀²························································································ ⒀（2分）

【評析】本題的關(guān)鍵在于平時對滑塊與滑板疊加問題的積淀，能快速準確分析相對運動的臨界條件、運動過程中出現(xiàn)速度相同的物體間摩擦力的轉(zhuǎn)型、快速準確運用牛頓運動定律及運動學(xué)公式計算各個物體的加速度、速度、位移及物體間相對位移。在處理問題的方法與技巧上要特別重視動量及能量守恒往往能快刀斬亂麻地找到初末狀態(tài)的速度，很好地回避中間過程，這也是兩大守恒的抽象思維能力的體現(xiàn)，在運用時要清醒的認識守恒的對象、過程與條件。另外本題后面的運動過程未設(shè)置問題，實際上A與C相對靜止后，將一起加速運動到0.6v₀，整個過程中B的運動最簡單，可考慮A與B最終相距多遠、整個過程中A、B與木板C因摩擦所產(chǎn)生的熱量之比、運動時間等問題。特別提醒：運動學(xué)公式、動量及能量中的矢量均以相對地面的物理量代入公式計算，相對位移用對地的位移差或功能關(guān)系解決，相對路程問題（具有重復(fù)劃痕的往返相對運動）要么分段、要么用功能關(guān)系解決。